Solution
= Solution
Under $P$, the first assumption yields $Q\to R$, while the second yields $Q$. Apply the <implication elimination rule> once more to obtain $R$, and discharge the three temporary assumptions:
|| Line
|| Context
|| Formula
|| Justification
| 1
| $\{1\}$
| $P\to(Q\to R)$
| Temporary assumption
| 2
| $\{1,2\}$
| $P\to Q$
| Temporary assumption
| 3
| $\{1,2,3\}$
| $P$
| Temporary assumption
| 4
| $\{1,2,3\}$
| $Q\to R$
| <implication elimination rule> $\to\text{-elim}$ on 1, 3
| 5
| $\{1,2,3\}$
| $Q$
| <implication elimination rule> $\to\text{-elim}$ on 2, 3
| 6
| $\{1,2,3\}$
| $R$
| <implication elimination rule> $\to\text{-elim}$ on 4, 5
| 7
| $\{1,2\}$
| $P\to R$
| <implication introduction rule> $\to\text{-intro}$ on 6; discharge 3
| 8
| $\{1\}$
| $(P\to Q)\to(P\to R)$
| <implication introduction rule> $\to\text{-intro}$ on 7; discharge 2
| 9
| $\varnothing$
| $(P\to(Q\to R))\to((P\to Q)\to(P\to R))$
| <implication introduction rule> $\to\text{-intro}$ on 8; discharge 1
The two uses of $P$ are permitted in ordinary <natural deduction>. Therefore
$$
\boxed{\vdash(P\to(Q\to R))\to((P\to Q)\to(P\to R)).}
$$